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Deriving 2D Motion Equations

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  1. Oct 7, 2015 #1
    1. The problem statement, all variables and given/known data
    For an upcoming test on 2D Motion, my Physics Teacher recommended that, instead of memorizing close to a dozen equations, he suggested we derive what we can from the main equations he showed us. So people in my class decided to split up deriving equations to bring them together later for everyone to use. I was told to use these two:
    y = (Vosin(theta))t - 1/2gt^2
    Vy = Vosin(theta) - gt

    So I was attempting to derive Vy from y, but found an issue.

    2. Relevant equations
    y = (Vosin(theta))t - 1/2gt^2
    Vy = (Vosin(theta)) - gt
    V = d/t

    3. The attempt at a solution
    To derive Vy, I simply did ((Vosin(theta))t - 1/2gt^2)) / t. This gave me (Vosin(theta)) - 1/2gt. This differs from the given equation for Vy, so I'm curious what I have to do to cancel out the 1/2.
     
  2. jcsd
  3. Oct 7, 2015 #2
    Do you know any calculus? That's the way to derive the equations.
     
  4. Oct 7, 2015 #3
    Yes, I am in calculus this year! I didn't even think to find d/dt. This makes things a lot easier, thanks!

    Using calculus, just to be sure I'm right, you can find

    y' = Vy
    Vy = 1 * (Vosin(theta)) - 1/2gt * 2, giving
    Vy = (Vosin(theta)) - gt

    That works out really well! Can that apply to a lot of other equations of motion (in Kinematics) as well?
     
  5. Oct 7, 2015 #4
    That's where kinematics equations come from!

    It all comes from the fact that [itex]a = \frac{d^2 x}{dt^2}[/itex] and the assumption that [itex]a[/itex] is constant.

    If you've learned about antidifferentiation (essentially the opposite of differentiation), then you can derive your equations from that equation alone. Since [itex]a = \frac{d v}{dt}[/itex], [itex]v = \int a \ dt = v_0 + a t[/itex]. And since [itex]v = \frac{d x}{dt}[/itex], [itex]x = \int v \ dt = x_0 + v_0 t + \frac{1}{2} a t^2 [/itex]. If you haven't learned about antidifferentiation, you can at least confirm that you can work backwards from those by taking derivatives to get back to [itex]\frac{d^2 x}{dt^2} = a[/itex].

    The equation that is not time-dependent is a bit trickier to derive, but that's essentially how the "usual" kinematics equations are derived. It all comes back to calculus--the mathematics of change.
     
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